Problem 83 - Applied Calculus: For Business, Economics, and the Social and Life Sciences

Question

TEMPERATURE CHANGE At \(t\) hours past midnight, the temperature \(T\left({ }^{\circ} \mathrm{C}\right)\) in a certain northern city is found to be changing at a rate given by \(T^{\prime}(t)=-0.02(t-7)(t-14) \quad{ }^{\circ}\) C/hour By how much does the temperature change between 8 A.M. and 8 P.M.?

Step-by-Step Solution

Answer

The temperature decreases by 8.53°C.

Step 1 - Understand the Problem

The problem gives the rate of temperature change as a function of time and asks for the total change in temperature over a specific period.

Step 2 - Determine the Time Interval

The interval is from 8 A.M. to 8 P.M. In terms of hours past midnight, 8 A.M. is 8 hours, and 8 P.M. is 20 hours.

Step 3 - Integrate the Temperature Change Rate

Integrate the rate of change function, \(T'(t) = -0.02(t-7)(t-14)\), from 8 to 20 to find the total change in temperature.

Step 4 - Set Up the Integral

The integral to solve is: \[ \int_{8}^{20} -0.02(t-7)(t-14) \, dt \]

Step 5 - Simplify the Integrand

Expand the integrand: \( -0.02(t-7)(t-14) = -0.02(t^2 - 21t + 98) \). Thus, the integral becomes: \[ -0.02 \bigg(\int_{8}^{20} (t^2 - 21t + 98) dt \bigg) \]

Step 6 - Integrate Term by Term

Integrate each term separately: \[ -0.02 \bigg( \left[ \frac{t^3}{3} - \frac{21t^2}{2} + 98t \right]_{8}^{20} \bigg) \]

Step 7 - Evaluate the Integral at the Bounds

Substitute the bounds into the antiderivative: \[ -0.02 \bigg( \left( \frac{20^3}{3} - \frac{21 \times 20^2}{2} + 98 \times 20 \right) - \left( \frac{8^3}{3} - \frac{21 \times 8^2}{2} + 98 \times 8 \right) \bigg) \]

Step 8 - Simplify the Expressions

Compute the values: \[ \frac{20^3}{3} = \frac{8000}{3}, \frac{21 \times 20^2}{2} = 4200, 98 \times 20 = 1960 \] for the upper bound, and \[ \frac{8^3}{3} = \frac{512}{3}, \frac{21 \times 8^2}{2} = 672, 98 \times 8 = 784 \] for the lower bound.

Step 9 - Calculate the Results

Combine the results: Upper bound is \( \frac{8000}{3} - 4200 + 1960 \) and lower bound is \( \frac{512}{3} - 672 + 784 \). Thus, the temperature change is: \[ -0.02 \bigg( \left( 2666.67 - 4200 + 1960 \right) - \left( 170.67 - 672 + 784 \right) \bigg) \]

Step 10 - Final Calculation

Subtract the lower bound from the upper bound and multiply by -0.02: \[ -0.02 \bigg( 426.67 \bigg) = -8.53 \]

Key Concepts

Rate of Change

In mathematics, the rate of change represents how a quantity changes relative to another variable, often time. When dealing with temperature change, we often use the derivative of the temperature function, denoted as \( T'(t) \), to understand how temperature varies over time. The exercise provides us with a specific rate of change function: \[ T'(t) = -0.02(t-7)(t-14) \] This equation shows how temperature changes hourly. By analyzing this, we can understand how rapidly the temperature increases or decreases over a given period.
In the context of the problem, the given rate of change allows us to calculate the total temperature change between specific times, which involves another powerful mathematical tool: integration.

Integration

Integration is a fundamental concept in calculus, helping us find the total accumulation of a quantity, such as distance, area, or in this case, temperature change. To determine how much the temperature changes between 8 A.M. and 8 P.M., we need to integrate the rate of change function over the given time interval. This means we calculate the definite integral of \( T'(t) \) from 8 to 20, which mathematically looks like this:
\[\begin{aligned} \ & \text{Integral} \ \int_{8}^{20} -0.02(t-7)(t-14) \, dt \end{aligned} \]
Following the steps of integration, we first expand the integrand and then integrate term by term:
\[\begin{aligned} -0.02 \bigg( \left[ \frac{t^{3}}{3} - \frac{21 t^{2}}{2} + 98 t \right]_{8}^{20}\bigg) \end{aligned} \]
Finally, by substituting the limits (20 and 8) and computing the values, we get the total change in temperature over the specified period. This process demonstrates the importance of integration in solving real-world problems involving cumulative changes.

Temperature Function

A temperature function describes how temperature evolves over time. In this problem, the derivative \( T'(t) \) represents the rate at which temperature changes. To find the actual change in temperature between two points in time, we use the integral of this derivative. By integrating the rate function \( T'(t) \), we can derive the total temperature change over our specified interval. This helps us capture the overall effect of the changing temperature from 8 A.M. to 8 P.M.
For this problem, after performing all necessary calculations, we found:
\[\begin{aligned} -0.02 \bigg( \left( \frac{8000}{3} - 4200 + 1960 \right) - \left( \frac{512}{3} - 672 + 784 \right) \bigg) \end{aligned}\right) \]
After evaluating and simplifying, the resulting temperature change is -8.53 degrees Celsius, indicating a decrease in temperature over the 12-hour period. Understanding and interpreting temperature functions are crucial for many applications, such as weather forecasting and climate studies.